Question

If $\omega (\ne 1)$ is cube root of unity satisfying $\frac{1}{a+\omega }+\frac{1}{b+\omega }+\frac{1}{c+\omega }=2{\omega }^2$ and $\frac{1}{a+{\omega }^2}+\frac{1}{b+{\omega }^2}+\frac{1}{c+{\omega }^2}=2\omega$, then the value of $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$ is :

• A. $2$
• B. $\omega$
• C. $-1$
• D. $-2$

Answer 1

The correct option is A $2$

We have $\omega (\ne 1)$ is cube root of unity satisfying $\frac{1}{a+\omega }+\frac{1}{b+\omega }+\frac{1}{c+\omega }=2{\omega }^2=\frac{2}{\omega }$ and
$\frac{1}{a+{\omega }^2}+\frac{1}{b+{\omega }^2}+\frac{1}{c+{\omega }^2}=2\omega =\frac{2}{{\omega }^2}$
$\therefore \omega$ and ${\omega }^2$ are roots of the equation
$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}=\frac{2}{x}\cdots \left(1\right)$

Simplifying the equation, we get

$\Rightarrow x\left[\right(x+a\left)\right(x+c)+(x+a\left)\right(x+b)+(x+b\left)\right(x+c\left)\right]=2(x+a)(x+b)(x+c)\Rightarrow x[3x^2+2(a+b+c)x+ab+bc+ca]=2[x^3+(a+b+c)x^2+(ab+bc+ca)x+abc]\Rightarrow x^3-(ab+bc+ca)x-2abc=0$
If the roots of the cubic equations are $\omega ,{\omega }^2,\gamma$, so from sum of the roots
$\omega +{\omega }^2+\gamma =0\Rightarrow \gamma =1$

Hence putting $x=1$ in $\left(1\right)$, we get
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2$